$\sinh \left(z^2\right)=\frac{15}{8}$
$$ \begin{aligned} & \text{Let } z=x+jy, \quad j^2 = -1 \\ & \frac{e^{z^2}-e^{-z^2}}{2}=\frac{15}{8} \\ & e^{z^2}-\frac{1}{e^{z^2}}=\frac{15}{4} \\ & A-\frac{1}{A}=\frac{15}{4} \\ & 4A^2-15A-4=0 \\ & (4A+1)(A-4)=0 \\ & A=4,-\frac{1}{4}, \text{ where } A = e^{z^2} \end{aligned} $$
$\mathbf{I}. e^{z^2} = 4$
$$\begin{aligned} &\text{from } z=x+j y \end{aligned}$$ $$ \begin{aligned} & \therefore z^2=x^2-y^2+j 2 x y \\ e^{\left(x^2-y^2\right)+2 j x y} & =4 \\ e^{\left(x^2-y^2\right)} e^{2 x y j} & =4 e^{j(0+2 n \pi)} \\ \therefore e^{x^2-y^2}=4 & , 2 x y=2 n \pi \end{aligned} $$
$$
\begin{aligned}
& x^2-y^2=\ln 4, \quad x y=n \pi \\
& x^2 y^2=n^2 \pi^2 \\
& \left(y^2+\ln 4\right) y^2=n^2 \pi^2 \\
& y^4+(\ln 4 )y^2=n^2 \pi^2 \\
& y^2=\frac{-\ln 4 \pm \sqrt{\ln ^2 4+4 n^2 \pi^2}}{2} \\
& \because y^2>0 ; y^2=\frac{\sqrt{\ln ^2 4+4 n^2 \pi^2}-\ln 4}{2} \\
& y^2=\sqrt{\ln ^2 2+n^2 \pi^2}-\ln 2 \\
& y= \pm \sqrt{ \sqrt{\ln ^2 2+n^2 \pi^2}-\ln 2 }\\
& x^2=y^2+\ln 4=\sqrt{\ln ^2 2+n^2 \pi^2}+\ln 2 \\
& x= \pm \sqrt{\sqrt{\ln ^2 2+n^2 \pi^2}+\ln 2} \\
&
\end{aligned}
$$
$\because x y=n\pi$
when $n=0 \rightarrow y=0$ and $x= \pm \sqrt{\ln 4} \rightarrow z= \pm \sqrt{\ln{4}}$
when $n>0 \rightarrow\{x>0, y>0]$ or $[x<0, y<0\} ; z= \pm \sqrt{\sqrt{\ln ^2 2+n^2 \pi^2}+\ln 2} \pm j \sqrt{\sqrt{\ln ^2 2+n^2 \pi^2}-\ln 2}$
when $n<0 \rightarrow[x<0, y>0]$ or $[x>0, y<0] ; z= \pm \sqrt{\sqrt{\ln ^2 2+n^2 \pi^2}+\ln 2} \mp j \sqrt{\sqrt{\ln ^2 2+n^2 \pi^2}-\ln 2}$
$\mathbb{II}. \quad e^{z^2} = -\dfrac{1}{4}$
$$ \begin{aligned} & e^{x^2-y^2} = \ln{\frac{1}{4}} \text{ , } e^{2 x y j}=\frac{1}{4} e^{j(\pi+2 n \pi)} \\ & e^{x^2-y^2} = \ln \left(\frac{1}{4}\right), \quad 2 x y=(2 n+1) \pi \\ & x^2-y^2=-\ln 4, x y=\left(n+\frac{1}{2}\right) \pi \\ & x^2=y^2-\ln 4 \\ & x^2 y^2=\left(n+\frac{1}{2}\right)^2 \pi^2 \\ & y^2\left(y^2-\ln 4\right)=\left(n+\frac{1}{2}\right)^2 \pi^2 \\ & y^4-(\ln 4 )y^2=\left(n+\frac{1}{2}\right)^2 \pi^2 \\ & y^2=\frac{\ln 4 \pm \sqrt{\ln ^2 4+4\left(n+\frac{1}{2}\right)^2 \pi^2}}{2} \\ \because y^2 \geqslant 0 \\ & y^2=\ln 2+\sqrt{\ln ^2 2+\left(0+\frac{1}{2}\right)^2 \pi^2} \\ & y= \pm \sqrt{\ln 2+\sqrt{\ln ^2 2+\left(n+\frac{1}{2}\right)^2 \pi^2}}\\ & x^2=y^2-\ln 4 \\ & =\sqrt{\ln ^2 2+\left(n+\frac{1}{2}\right)^2 \pi^2}-\ln 2 \\ & x= \pm \sqrt{\sqrt{\ln ^2 2+\left(n+\frac{1}{2}\right)^2 \pi^2}-\ln 2} \\ & \begin{array}{l} \text { when } n \geqslant 0 \rightarrow x y \geqslant 0 \\ z = x+j y= \pm \sqrt{\sqrt{\ln ^2 2+\left(n+\frac{1}{2}\right)^2 \pi^2}-\ln 2} \pm j \sqrt{\sqrt{\ln ^2 2+\left(n+\frac{1}{2}\right)^2 \pi^2}+\ln 2} \end{array} \\ & \text { when } n<0 \rightarrow x y<0 \\ & z=x+j y= \pm \sqrt{\sqrt{\ln ^2 2+\left(n+\frac{1}{2}\right)^2 \pi^2-}\ln 2} \mp j \sqrt{\sqrt{\ln ^2 2+\left(n+\frac{1}{2}\right)^2 \pi^2}+\ln 2} \\ & \end{aligned} $$
Is there any point I do wrong? I’m not sure. I wonder if there’s other method easier than I do.
